herdoc和PHP的问题

herdoc和PHP的问题,php,mysql,Php,Mysql,我正在关注Wrox的《开始PHP、Apache、MySQL web开发》一书。我一直在一字不差地关注它,出于某种原因,我对代码有一个问题。编辑说我的代码中没有错误。但当我运行时,它会给我以下信息: “您的SQL语法有错误;请查看与您的MySQL服务器版本相对应的手册,以了解在第3行的“”附近使用的正确语法”以下是代码 <?php //take in the id of a director and return his/her full name function get_director

我正在关注Wrox的《开始PHP、Apache、MySQL web开发》一书。我一直在一字不差地关注它,出于某种原因,我对代码有一个问题。编辑说我的代码中没有错误。但当我运行时,它会给我以下信息: “您的SQL语法有错误;请查看与您的MySQL服务器版本相对应的手册,以了解在第3行的“”附近使用的正确语法”以下是代码

<?php
//take in the id of a director and return his/her full name
function get_director() {

global $db;

$query = 'SELECT people_fullname
          FROM people
          WHERE  people_id = ' . $director_id;
$result = mysql_query($query, $db) or die(mysql_error($db));

$row = mysql_fetch_assoc($result);
extract($row);

return $people_fullname;
}

//take in the id of a lead actor and return his/her full name
function get_leadactor($leadactor_id) {

global $db;

$query = 'SELECT people_fullname
          FROM people
          WHERE people_id = ' . $leadactor_id;
$result = mysql_query($query, $db) or die (mysql_error($db));
$extract($row);

return $people_fullname;
}

// take in the id of a movie type 
// and return the meaningful textual description
function get_movietype($type_id) {

global $db;

$query = 'SELECT movietype_label
          FROM movietype
          WHERE movietype_id = ' . $type_id;
          $result = mysql_query($query, $db) or die (mysql_error($db));

          $row = mysql_fetch_assoc($result);
          extract($row);

          return $movietype_label;
 }
   // conect to MySQL
   $db = mysql_connect('localhost', 'root', 'root') or 
   die ('unable to connect. Check your parameters');

    // make sure you are yousing the right database
    mysql_select_db('moviesite', $db) or die(mysql_error($db) );

    // retrieve information
    $query = 'SELECT movie_name, movie_year, movie_director, movie_leadactor, movie_type
           FROM movie
           ORDER BY movie_name ASC, movie_year DESC';
    $result = mysql_query($query, $db) or die ($mysql_error($db) );

    // determine number of rows in returned result
    $num_movies = mysql_num_rows($result);


    $table = <<<ENDHTML
    <div style ="text-align: center;">
    <h2>Movie Review Database</h2>
      <table border="1" cellpadding="2" cellspacing="2" style="width: 70%; margin-left: auto;     margin-right:auto;">
     <tr>
       <th>Movie Title</th>
       <th>Year of the release</th>
       <th>Movie Director</th>
       <th>Movie Lead Actor</th>
       <th>Movie Type</th>
     </tr>
     ENDHTML;

      //loop throught the results
      while ($row = mysql_fetch_assoc($result) ) {
      extract($row);
      $director = get_director($movie_director);
      $leadactor = get_leadactor($movie_leadactor);
      $movietype = get_movietype($movie_type);

      $table .= <<<ENDHTML
      <tr>
         <td>$movie_name</td>
         <td>$movie_year</td>
         <td>$director</td>
         <td>$leadactor</td>
         <td>$movietype</td>
     </tr>
     ENDHTML;

  }

  $table .= <<<ENDHTML
  </table>
  <p>$num_movies Movies</p>
  </div>
  ENDHTML;

  echo $table
  ?>

在那之后,我试着复制和粘贴确切的代码,认为可能是我做错了什么 这是下面的代码:

 <?php
 // take in the id of a director and return his/her full name
 function get_director($director_id) {
 global $db;
 $query = 'SELECT
 people_fullname
 FROM people
 WHERE
 people_id = ' . $director_id;
 $result = mysql_query($query, $db) or die(mysql_error($db));
 $row = mysql_fetch_assoc($result);
 extract($row);
 return $people_fullname;
 }
 // take in the id of a lead actor and return his/her full name
 function get_leadactor($leadactor_id) {
 global $db;
 $query = 'SELECT
 FROM
 people WHERE
 people_id = ' . $leadactor_id;
 $result = mysql_query($query, $db) or die(mysql_error($db));
 $row = mysql_fetch_assoc($result);
 extract($row);
 return $people_fullname;
 }
 // take in the id of a movie type and return the meaningful textual
 // description
 function get_movietype($type_id) {
 global $db;
 $query = 'SELECT
 movietype_label
 FROM
 movietype
 WHERE
 movietype_id = ' . $type_id;
 $result = mysql_query($query, $db) or die(mysql_error($db));
 $row = mysql_fetch_assoc($result);
 extract($row);
 return $movietype_label;
 }

 //connect to MySQL
 $db = mysql_connect('localhost', 'root', 'root') or
 die ('Unable to connect. Check your connection parameters.');
 // make sure you’re using the right database
 mysql_select_db('moviesite', $db) or die(mysql_error($db));
 // retrieve information
 $query = 'SELECT
 movie_name, movie_year, movie_director, movie_leadactor,
 movie_type
 FROM
 movie
 ORDER BY
 movie_name ASC,
 movie_year DESC';
 $result = mysql_query($query, $db) or die(mysql_error($db));
 // determine number of rows in returned result
 $num_movies = mysql_num_rows($result);

 $table = <<<ENDHTML

 <div style="text-align: center;">
 <h2>Movie Review Database</h2>
 <table border="1" cellpadding="2" cellspacing="2"
 style="width: 70%; margin-left: auto; margin-right: auto;">
 <tr>
  <th>Movie Title</th>
  <th>Year of Release</th>
  <th>Movie Director</th>
  <th>Movie Lead Actor</th>
  <th>Movie Type</th>
 </tr>

 ENDHTML;
  // loop through the results
  while ($row = mysql_fetch_assoc($result)) {
  extract($row);
  $director = get_director($movie_director);
  $leadactor = get_leadactor($movie_leadactor);
  $movietype = get_movietype($movie_type);
  $table .= <<<ENDHTML
  <tr>
   <td>$movie_name</td>
   <td>$movie_year</td>
   <td>$director</td>
   <td>$leadactor</td>
   <td>$movietype</td>
  </tr>
   ENDHTML;
 }
 $table .= <<<ENDHTML
 </table>
 <p>$num_movies Movies</p>
 </div>
 ENDHTML;

 echo $table;
 ?> 

当我这次运行代码时,表格标题显示,但部分代码也显示在浏览器上。看起来是这样的:
ENDHTML;//在(=mysql_fetch_assoc(资源id#3)){extract();=get_director();=get_leadactor();=get_movietype()时循环遍历结果;.=当你结束一个heredoc时,你不能在行首放任何字符。在你的代码中,有一些heredoc前面有空格或制表符。删除空格并将你的heredoc放在行首


结束语句必须位于字符串的第一个位置:

ENDHTML;    // works
  ENDHTML;  // won’t work

请注意,它们已不再维护,而是。请改为了解,并使用or。您还希望在heredoc的结束标识符之前不能有任何空格或制表符。您只需在必要的表中进行一次查询,不要在循环中查询其他信息。回显您的
$query
,您将看到pro有问题。很可能是这些行的报价问题
WHERE people\u id=”。$director\u id;
spaces ENDHTML;
如前所述。删除
ENDHTML;
之前和之后的空格。如果设置了,则会给您一个错误。谢谢大家。所以很清楚,没有人会在他们的网站上使用这种方法吗?